Solving Linear Equations Y(4) Y + 7y + 6y 8y = 0

SOLVING LINEAR EQUATIONS (4)6+7+68=0 (0)=2,(0)=8,(0)=14,(0)=62 The characteristic polynomial is: 463+72+68=0 The constant term ir =. From the rational root theorem, the possible roots are 1,2,4, 8. Then �1(1)=16+7+68=0 so by the remainder theorem, =1 is a lot. Dividing �1(1) by 1 using synthetic division, we obtain: 1 -6 7 6 -8 1 1 -5 2 8 1 -5 2 8 0 Let �2()=352+2+8. checking for =1,�2(1)=15+2+80, so we now discard =1, Moving on to the next wot, =1, we obtain 1 -5 2 -8 -1 -1 6 8 -1 -6 8 0 Let �3()=26+8=0. Checking for =1,�3(1)=16+8+0.50 we finally discard =1. Since the function is quadratic, it will be inefficient to use the Rational root. By factoring, we obtain �4()=(4)(2) Equating both factors to zero, we get the last two roots =2 and =4. Therefore, the roots are: 1,1,2,4. Hence, the general solution is given by: =�1�+�2�+�3�2+�4�4 The first, second and third derivatives of are:=�1�+�2�+2�3�2+4�4�4=�1�+�2�+4�3�2+16�4�4=�1�+�2�+8�3�2+64�4� Plugging in the initial values to anode its derivatives, we obtain the syrup of equations: �1+�2+�3+�4=2 �1+�2+2�3+4�4=8�1+�2+4�3+16�4=14 �1+�2+8�3+64�4=62 The coefficient matrix and vector for this system are =[111111241141611864],�=[281462] Thus, �1=|21118124141416621864||111111241141411844|=360180=2�2=|12111824114416162864|||=720180=43=|11211184111416116264|11=180180=1 �4=|111211281141411862|||=180180=1 Writing the particular solution, �=��+��